Uche pumps gasoline at a rate of $18\,\dfrac{\text{L}}{\text{min}}$. What is Uche's pumping rate in $\dfrac{\text{mL}}{\text{s}}$ ?
Explanation: We will convert $18\,\dfrac{\text{L}}{\text{min}}$ to a rate in $\dfrac{\text{mL}}{\text{s}}$ using the following conversion rates: There are $1000\text{ mL}$ per $1\text{ L}$. There is $1\text{ min}$ per $60\text{ s}$. $\begin{aligned} &\phantom{=}\dfrac{18\text{ L}}{1\text{ min}}\cdot\dfrac{1000\text{ mL}}{1\text{ L}}\cdot\dfrac{1\text{ min}}{60 \text{ s}} \\\\ &=\dfrac{18 \cdot 1000 \cdot 1 \cdot \cancel{\text{L}} \cdot \text{mL} \cdot \cancel{\text{min}}}{1 \cdot 1 \cdot 60 \cdot \cancel{\text{min}} \cdot \cancel{\text{L}} \cdot \text{s}} \\\\ &=\dfrac{18{,}000}{60}\,\dfrac{\text{mL}}{\text{s}} \\\\ &=300\,\dfrac{\text{mL}}{\text{s}} \end{aligned}$ In conclusion, Uche's rate in $\dfrac{\text{mL}}{\text{s}}$ is: $300\,\dfrac{\text{mL}}{\text{s}}$